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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS
+
+// EXAMPLE : 8.6 :
+// Page number 626
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA = 50.0 // Alternator rating(MVA)
+kV = 11.0 // Alternator voltage(kV)
+X = 2.0 // Synchronous reactance per phase(ohm)
+R = 0.7 // Resistance per phase(ohm)
+R_n = 5.0 // Resistance through which alternator is earthed(ohm)
+ofb = 25.0 // Out-of-balance current(%)
+
+// Calculations
+I_fl = MVA*1000/(3**0.5*kV) // Full load current(A)
+I_ofb = ofb/100*I_fl // Out-of-balance current(A)
+x = R_n/((kV*1000/(3**0.5*100*I_ofb))-(R/100)) // Portion of winding unprotected(%)
+
+// Results
+disp("PART III - EXAMPLE : 8.6 : SOLUTION :-")
+printf("\nPortion of winding unprotected, x = %.f percent", x)